Question

In ∆ABC, seg BD bisects ∠ABC. If AB = x, BC = x + 5, AD = x – 2, DC = x + 2, then find the value of x.

Answer 1

In △ABC, ∠ABD = ∠DBC
$$\begin{gathered} \frac{\mathrm{AD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{CB}}\left(\mathrm{By}\mathrm{angle}\mathrm{bisector}\mathrm{theorem}\right) \\ \Rightarrow \frac{x-2}{x+2}=\frac{x}{x+5} \\ \Rightarrow x^2+3x-10=x^2+2x \end{gathered}$$
$$\begin{gathered} \Rightarrow 3x-2x=10 \\ \Rightarrow x=10 \end{gathered}$$